Question: $ \left(\dfrac{64}{9}\right)^{-\frac{1}{2}}$
$= \left(\dfrac{9}{64}\right)^{\frac{1}{2}}$ Figure out what goes in the blank: $\Big(? \Big)^{2}=\dfrac{9}{64}$ Figure out what goes in the blank: $\Big({\dfrac{3}{8}}\Big)^{2}=\dfrac{9}{64}$ So $\left(\dfrac{64}{9}\right)^{-\frac{1}{2}}=\left(\dfrac{9}{64}\right)^{\frac{1}{2}}=\dfrac{3}{8}$